SOLUTION: Hi a solid metal sphere of radius 21cm is melted and recast into 2 cylinders each with end radius of 14 cm, but with one cylinder twice the length of the other.assuming no loss

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Question 1011098: Hi
a solid metal sphere of radius 21cm is melted and recast into 2 cylinders each with end radius of 14 cm, but with one cylinder twice the length of the other.assuming no loss of metal what will be the length of each cylinder.
How does the total surface area of the 2 cylinder s compare with the surface area of the original sphere.
thanks
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
R, known radius of the sphere
r, known radius of each cylinder
h, length of small cylinder
2h, length of large cylinder

First find h.
Sum of volumes is the original sphere's volume.

highlight_green%28%284%2F3%29pi%2AR%5E3=h%2Api%2Ar%5E2%2B2h%2Api%2Ar%5E2%29
h%28pi%2Ar%5E2%2B2pi%2Ar%5E2%29=%284%2F3%29pi%2AR%5E3
h%2Api%2A%28r%5E2%2B2r%5E2%29=%284%2F3%29pi%2AR%5E3
h=%28%284%2F3%29R%5E3%29%2F%28r%5E2%2B2r%5E2%29
highlight%28h=%284R%5E3%29%2F%283r%5E2%2B6r%5E2%29%29

Evaluate h using system%28R=21%2Cr=14%29.

Use formulas for surface areas to answer the second question.
If X is radius in general, and L is length in general:
Sphere: 4pi%2AX%5E2
Cylinder: %282pi%2AX%29%2AL%2B2%2Api%2AX%5E2